Equation of circle though points of intersection of conics

Question

The possible solution as per hint offered in my textbook is:

Hence the final equation of circle would be 13x²+13y²+18x+8y-59=0.

But I cannot figure out why the slope of line UV should be 1.

Answer 1

A simple sketch reveals that the tangent points are $(-2, 0)$ and $(0, -2)$, hence $M, N$ are $(-1,0)$ and $(0, -1)$, and the line $MN$ has the equation $ x + y = -1 $. Intersecting this line with the given ellipse, results in a quadartic equation in $x$

$ \dfrac{x^2}{9} + \dfrac{(x + 1)^2}{4} = 1 $

Multiply through by $36$

$ 4 x^2 + 9 (x + 1)^2 = 36 $

which leads to

$ 13 x^2 + 18 x - 27 = 0 $

which has the following roots

$x = \dfrac{ - 9 \pm 12 \sqrt{3} }{13} $

Hence, the center of the circle $RS$ is at

$ C = ( - \dfrac{9}{13}, -\dfrac{4}{13} ) $

The radius is $ \sqrt{2} \times \dfrac{12 \sqrt{3}}{13} = \dfrac{12 \sqrt{6}}{13} $

Therefore, the equation of the circle containing $R, S, U, V$ is

$( x + \dfrac{9}{13} )^2 + (y + \dfrac{4}{13})^2 = \left(\dfrac{12}{13}\right)^2 \times 6 $

Multiplying through by $(13)^2$

$ (13 x + 9)^2 + ( 13 y + 4 )^2 = 864 $